Science One Math March 1, 2017 Applications of Integration - PowerPoint PPT Presentation

Science One Math March 1, 2017

Applications of Integration • Computing areas • Computing changes • Computing volumes • Computing probabilities • Locating centre of mass of an object • Work done by a non constant force

Application of Integration: Work What is work?

Application of Integration: Work What is work? Work is change in energy when a force causes displacement. Basic formula: work = force x displacement valid if force is constant !

Work done by a non-constant force If F is constant, 𝑋 = 𝐺 $ 𝑦 If F is not constant , add up small amounts of work over short distances Δ𝑦 Key ideas: split displacement into small segments Δ𝑦 , assume F to be constant over Δ𝑦 find Δ𝑋 work done by F over Δ𝑦

Strategy for computing work done by varying F • divide displacement into n segments of length Δ𝑦 • assume force is a constant over each segment (possibly different for each ∗ ) segment), force resulting in k -th segment of displacement = 𝐺(𝑦 ( ∗ Δ𝑦 • compute work done by constant force over short displacement Δ𝑋 = 𝐺 𝑦 ( - ∗ )Δ𝑦 • add up small amounts of work, we get a Riemann Sum ∑ 𝐺(𝑦 ( (./ • take the limit for 𝑜 → ∞ , we get a definite integral 9 - ∗ )Δ𝑦 = ∫ 𝐺 𝑦 𝑒𝑦 -→ 6 ∑ 𝑋 = lim 𝐺(𝑦 ( (./ :

Fundamental Theorem of Calculus is fundamental in Physics too! FTC tells us how to compute a definite integral 9 𝑋 = ∫ 𝐺 𝑦 𝑒𝑦 = 𝑉 𝑐 − 𝑉(𝑏) : where 𝑉 𝑦 = ∫ 𝐺 𝑦 𝑒𝑦 + 𝐷 . We call 𝑉 𝑦 the potential energy . AB FTC also tells us 𝐺 𝑦 = AC (for an external force exerted on object). AB Note for a force exerted by the potential energy we have 𝐺 𝑦 = − AC

Stretching and compressing a spring Hooke’s law : force required to keep a spring compressed or stretched a distance x is proportional to x.

Stretching and compressing a spring Hooke’s law : force required to keep a spring compressed or stretched a distance x is proportional to x. 1) Force F exerted on spring is in the same direction as displacement (both positive when stretching, both negative when compressing). Work done to stretch/compress a spring by x is C C = ∫ 𝑙𝑢 𝑒𝑢 = / G 𝑙 𝑦 G 𝑋 = ∫ 𝐺 𝑢 𝑒𝑢 E E

Stretching and compressing a spring Hooke’s law : force required to keep a spring compressed or stretched a distance x is proportional to x. 1) Force F exerted on spring is in the same direction as displacement (both positive when stretching, both negative when compressing). Work done to stretch/compress a spring by x is C C = ∫ 𝑙𝑢 𝑒𝑢 = / G 𝑙 𝑦 G 𝑋 = ∫ 𝐺 𝑢 𝑒𝑢 E E 2) Force 𝐺 H done by spring is opposite to displacement Work done by a spring compressed by x, C 𝑋 = ∫ −𝑙𝑢 𝑒𝑢 = − / G 𝑙 𝑦 G E

Pulling up a heavy rope A 10-m long rope of density 2 kg/m is hanging from a wall which is 5 m high (so 5 m of rope runs down the length of the wall and the remaining 5 m is coiled at the bottom of the wall). How much work (in J) is required to pull the rope to the top of the wall? Let g be the acceleration due to gravity. /E A. 𝑋 = 𝑕 ∫ 2𝑧𝑒𝑧 E Q B. 𝑋 = 50𝑕 + 𝑕 ∫ 2𝑧𝑒𝑧 E /E C. 𝑋 = 50𝑕 + 𝑕 ∫ 2(10 − 𝑧)𝑒𝑧 E Q D. 𝑋 = 50𝑕 + 𝑕 ∫ 2(5 − 𝑧)𝑧𝑒𝑧 E Q E. 𝑋 = 50𝑕 + 𝑕 ∫ 2(5 − 𝑧)𝑒𝑧 E

Pulling up a heavy rope A 10-m long rope of density 2 kg/m is hanging from a wall which is 5 m high (so 5 m of rope runs down the length of the wall and the remaining 5 m is coiled at the bottom of the wall). How much work (in J) is required to pull the rope to the top of the wall? Let g be the acceleration due to gravity. Force exerted = mg but mass depends of length of rope hanging To pull up the first 5 m of rope we apply constant force of 2x5xg J To pull up the second 5 m of rope we apply a decreasing force because the mass of the rope is decreasing as more rope is pulled up Q 𝑋 = (50𝑕) + ∫ 2 5 − 𝑧 𝑕 𝑒𝑧 E force displacement

What if different “parts” of an “object” undergo different displacements? H Strategy: • slice object into n layers of thickness Δ𝑧 Δy y • assume force on each layer is constant (possibly different for each layer) • find work done on each layer by 0 multiplying force by displacement undergone by layer • add up all amounts of work.

What if different “parts” of an object undergo different displacements? H Strategy: • slice object into n layers of thickness Δ𝑧 Δy y • assume force on each layer is constant (possibly different for each layer) • find work done on each layer by 0 multiplying force by displacement 𝐺 j:klm = (𝑥𝑓𝑗𝑕ℎ𝑢) j:klm undergone by layer Δ𝑋 = (Δ𝑛 j:klm 𝑕) $ 𝑒𝑗𝑡𝑞𝑚𝑏𝑑𝑓𝑛𝑓𝑜𝑢 • add up all amounts of work. Δ𝑛 j:klm = 𝑒𝑓𝑜𝑡𝑗𝑢𝑧 ∆𝑊 j:klm , Δ𝑊 j:klm = 𝐵 j:klm Δ𝑧 | total work 𝑋 = ∫ 𝜍 𝐵 j:klm 𝑧 𝑕 𝑒𝑧 E

Building a pile of sand How much work must be done in producing a conical heap of sand of base radius R and height H? The specific weight of sand is 𝜍 . You may assume that all the sand is taken from the surface of the earth (that is, from height 0). Work to lift a layer of mass m up a height y from the ground force displacement Δ𝑋 = (𝑛 j:klm $ 𝑕) $ 𝑧 𝑛 j:klm = 𝜍 H:-A $ 𝑊 j:klm G € j:klm = 𝜌 𝑠 G Δ𝑧 = 𝜌 𝑆 − 𝑊 | 𝑧 Δ𝑧 G | 𝑕 𝜌 𝑆 − 𝑆 𝑋 = • 𝜍 H:-A 𝐼 𝑧 𝑧 𝑒𝑧 E

Digging a hole Consider two workers digging a hole. How deep should the first worker dig so that each does the same amount of work? Let the weight density of the dirt be the constant 𝜍 , the depth of the hole is D, and the cross- sectional area of the hole is A (so we assume that the hole does not get any wider or narrower as the workers dig). A. The first worker should dig to a depth 𝐸/2 B. The first worker should dig to a depth 𝐸/4 † C. The first worker should dig to a depth G D. The first worker should dig to a depth 2 𝐸

Digging a hole Consider two workers digging a hole. How deep should the first worker dig so that each does the same amount of work? Let the weight density of the dirt be the constant 𝜍 , the depth of the hole is D, and the cross- sectional area of the hole is A (so we assume that the hole does not get any wider or narrower as the workers dig). † 𝑧 𝑒𝑧 = 𝜍𝐵𝑕 𝐸 G 𝑋 ‡ˆ‡ = • 𝜍 𝐵 𝑕 2 E Let x be the depth of the hole the first worker digs, then we must solve the following equation for x, † ‰ C / † ∫ 𝜍 𝐵 𝑕 𝑧𝑒𝑧 = G 𝜍𝐵𝑕 G , which yields 𝑦 = G . E

Pumping out fluid from a tank A cylindrical tank with a length of L m and a radius of R m is on its side and half-full of gasoline. How much work is done to empty the tank through an outlet pipe at the top of the tank? Let ρ be density of gasoline, and A be the cross-sectional area of a layer at height y, G€ A. ∫ 𝜍 𝑕 𝐵 2𝑆 − 𝑧 𝑒𝑧 E G€ B. ∫ 𝜍 𝑕 𝐵 𝑆 − 𝑧 𝑒𝑧 E € C. ∫ 𝜍 𝑕 𝐵 2𝑆 − 𝑧 𝑒𝑧 E € D. ∫ 𝜍 𝑕 𝐵 𝑧 𝑒𝑧 E € E. ∫ 𝜍 𝑕 𝐵 𝑒𝑧 E

Pumping out fluid from a tank A cylindrical tank with a length of L m and a radius of R m is on its side and half-full of gasoline. How much work is done to empty the tank through an outlet pipe at the top of the tank? Let ρ be density of gasoline, and A be the cross-sectional area of a layer at height y, G€ A. ∫ 𝜍 𝑕 𝐵 2𝑆 − 𝑧 𝑒𝑧 E G€ B. ∫ 𝜍 𝑕 𝐵 𝑆 − 𝑧 𝑒𝑧 E € C. ∫ 𝜍 𝑕 𝐵 2𝑆 − 𝑧 𝑒𝑧 E € D. ∫ 𝜍 𝑕 𝐵 𝑧 𝑒𝑧 E € E. ∫ 𝜍 𝑕 𝐵 𝑒𝑧 E